#include <bits/stdc++.h>
#define rep(i,n) for(ll i = 0; i < (n); ++i)
#define drep(i,n) for(ll i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (ll i = s; i < t; ++i)
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
#define yn {puts("Yes");}else{puts("No");}

ll mergecount(vector<int> &a) {
    ll count = 0;
    int n = a.size();
    if (n > 1) {
        vector<int> b(a.begin(), a.begin() + n/2);
        vector<int> c(a.begin() + n/2, a.end());
        count += mergecount(b);
        count += mergecount(c);
        for (int i = 0, j = 0, k = 0; i < n; ++i)
        if (k == c.size())       a[i] = b[j++];
        else if (j == b.size())  a[i] = c[k++];
        else if (b[j] <= c[k])   a[i] = b[j++];
        else                   { a[i] = c[k++]; count += n/2 - j; }
    }
    return count;
}

const int MAX = 510000;
const int MOD = 998244353;
long long fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

long long FINV(int n){
    if (n < 0) return 0;
    return finv[n];
}

// ax + by = gcd(a, b) となるような (x, y) を求める
// a と b は互いに素として ax + by = 1 となる (x, y) を求める
long long extGCD(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    long long d = extGCD(b, a%b, y, x); // 再帰
    y -= a / b * x;
    return d;
}


// 負の数に対応した mod
inline long long mod(long long a, long long m) {
    return (a % m + m) % m;
}


// 逆元計算 (a と m が互いに素であることが必要)
long long modinv(long long a, long long m) {
    long long x, y;
    extGCD(a, m, x, y);
    return mod(x, m); // x % m だが、x が負かもしれないので
}

int main() {
    COMinit();
    ll n; cin >> n;
    vector<int> a(n);
    rep(i,n) cin >> a[i];
    ll ten = mergecount(a);
    ten %= MOD; 
    
    map<int,ll> mp;
    rep(i,n)mp[a[i]]++;
    ll diff = 0;
    rep(i,n){
        diff += (n-mp[a[i]]);
    }
    diff /= 2;
    // cout << "diff = " << diff << endl;

    ll mul = 1;
    srep(i,1,n+1){
        mul *= COM(n,i);
        mul %= MOD;
    }
    ll all = mul;
    ll sum2 = 0;
    srep(i,1,n+1){
        sum2 += (ll)i * (((n*(n+1)/2) - i*(i+1)/2)%MOD) % MOD;
        sum2 %= MOD;
    }
    all = all * sum2 % MOD;
    // all *= (n*(n+1)%MOD*((n*(n+1)-2)%MOD)%MOD*modinv(8,MOD)%MOD)%MOD;

    // cout << all << endl;

    ll mine = mul * modinv(n*n,MOD) % MOD;
    ll sum = 0;

    srep(i,1,n+1){
        ll tmp = n*(n+1)/2 - i*(i+1)/2;
        sum += tmp;
        sum %= MOD;
    }
    mine = mine * sum % MOD * n % MOD;

    // cout << mine << endl;

    // ll ans1 = (all + MOD - mine) % MOD * diff % MOD * modinv(n*(n-1),MOD) % MOD;
    ll ans1 = all % MOD * diff % MOD * modinv(n*n,MOD) % MOD;

    ll sum3 = 0;
    ll mine2 = mul * modinv(n*(n-1),MOD) % MOD;
    srep(i,2,n+1){
        sum3 += i * (i - 1);
        sum3 %= MOD;
    }
    mine2 = mine2 * sum3 % MOD;
    ll ans2 = mine2 * ten % MOD;
    ll ans = (ans1 + ans2) % MOD;

    // cout << ans1 << ' ' << ans2 << endl;
    cout << ans << endl;
    return 0;
}